I'm trying to expand a simple fibonacci function, and I need to use the values for each term more than once. So, I figured I'd use let to hold onto the values. But, I'm not getting what I think I should out of the function.
Here is the original fib function:
(define (fib n)
(if (< n 2)
n
(+ (fib (- n 1)) (fib (- n 2)))))
Here is my attempt at doing the same thing, but with let:
(define (fib-with-let n)
(if (< n 2)
0
(let ((f1 (fib-with-let (- n 1)))
(f2 (fib-with-let (- n 2))))
(+ f1 f2))))
Results:
> (fib 10)
55
> (fib-with-let 10)
0
Thanks!
You made a typo:
(if (< n 2)
0
...)
You mean n.
You mistyped your base case. In the first version you had:
(if (< n 2)
n
But then in your latter version you wrote:
(if (< n 2)
0
So just change 0 to n.
Your let is not really doing anything. You are still doing all of the extra calculations. Just because you define f1 as (fib-with-let (- n 1)) doesn't mean you won't compute the fib of n-1 again. f2 does not use f1. If you wanted f2 to see f1 you would use let*. However, even this is not really what you want.
As evidence of this, here are the running times for fib(35) and fib-with-let(35):
(time (fib 35))
cpu time: 6824 real time: 6880 gc time: 0
(time (fib-with-let 35))
cpu time: 6779 real time: 6862 gc time: 0
What you really want to do to avoid extra computations is use dynamic programming and recurse in a bottom-up fashion.
What you want is the following code:
(define (dynprog-fib n)
(if (< n 2)
n
(dynprog-fib-helper 1 1 2 n)))
(define (dynprog-fib-helper n1 n2 current target)
(if (= current target)
n2
(dynprog-fib-helper n2 (+ n1 n2) (add1 current) target)))
(time (dynprog-fib 35))
cpu time: 0 real time: 0 gc time: 0
(time (dynprog-fib 150000))
cpu time: 2336 real time: 2471 gc time: 644
As you can see, you can do the first 150,000 fibs in a third of the time the naive approach takes.
Since it looks like you are confused about what let does let me illustrate better:
When you say:
(let ((a 1)
(b 2))
(+ a b))
What you are saying is, let a be 1, and b be 2, add them together.
If you instead said:
(let ((a 1)
(b (+ a 1))
(+ a b))
Can you guess what you'd get? Not 3. It would be blow up with expand: unbound identifier in module in: a
In simple let, your assignments cannot see each other.
If you wanted to write the above you would have to use let*:
(let* ((a 1)
(b (+ a 1))
(+ a b))
That would give you the 3 you expect. let* essentially expands to:
(let ((a 1))
(let ((b (+ a 1)))
(+ a b)))
What you thought you were doing with the lets is called memoization. It's a technique where you store intermediate values so you don't have to repeat them. Let, however, does not do that for you.
Although your problem is a typo in your fib-with-let function, in its simplest form, let is "syntatic-sugar" for an anonymous lambda followed by the arguments that are then evaluated and passed to the lamba, which is then evaluated and a final value returned. So
(let ((f1 (fib-with-let (- n 1)))
(f2 (fib-with-let (- n 2))))
(+ f1 f2))
would be re-written without let to look like
((lambda (f1 f2) (+ f1 f2))(fib-with-let (- n 1))(fib-with-let (- n 2)))
Related
I am writing the square of sums in racket/scheme recursively. The code sums the numbers right, but it doesn't square it right. I don't know what I am doing wrong. If I pass 10, it should be 3025.
(define (squareOfSums n)
(if (= n 0)
0
(expt (+ n (squareOfSums (- n 1))) 2)))
You should do the squaring only once, at the end of the recursion. Currently, your code squares at every iteration. One way to solve this problem would be to separate the sum part into a helper procedure, and square the result of calling it. Like this:
(define (squareOfSums n)
(define (sum n)
(if (= n 0)
0
(+ n (sum (- n 1)))))
(sqr (sum n)))
Also, did you know that there's a formula to add all natural numbers up to n? This is a nicer solution, with no recursion needed:
(define (squareOfSums n)
(sqr (/ (* n (+ n 1)) 2)))
Either way, it works as expected:
(squareOfSums 10)
=> 3025
Here's a version which I think is idiomatic but which I hope no-one who knows any maths would write:
(define (square-of-sums n)
(let loop ([m n] [sum 0])
(if (> m 0)
(loop (- m 1) (+ sum m))
(* sum sum))))
Here's the version someone who knows some maths would write:
(define (square-of-sums n)
(expt (/ (* n (+ n 1)) 2) 2))
I wish people would not ask homework questions with well-known closed-form solutions: it's actively encouraging people to program badly.
If you start out with your function by writing out some examples, it will be easier to visualize how your function will work.
Here are three examples:
(check-expect (SquareOfSums 0) 0)
(check-expect (SquareOfSums 2) (sqr (+ 2 1))) ;9
(check-expect (SquareOfSums 10) (sqr (+ 10 9 8 7 6 5 4 3 2 1))) ;3025
As we can see clearly, there are two operators we are using, which should indicate that we need to use some sort of helper function to help us out.
We can start with out main function squareOfSums:
(define (squareOfSums n)
(sqr (sum n)))
Now, we have to create the helper function.
The amount of times that you use the addition operator depends on the number that you use. Because of this reason, we're going to have to use natural recursion.
The use of natural recursion requires some sort of base case in order for the function to 'end' somewhere. In this case, this is the value 0.
Now that we have identified the base case, we can create our helper function with little issue:
(define (sum n)
(if (= 0 n)
0
(+ n (sum (sub1 n)))))
I'll illustrate what I want to do using Python (I want to write this in Clojure). I have this function:
def f(n):
s=0
for d in range(1,n+1):
s+=d*(n//d)
return(s)
Which is basically looping from d=1 to n inclusive, and summing up the values of d times the floor of n/d.
In Clojure I want to make this a recursive function. Python equivalent:
def f(d, n):
if d == 0: return 0
else: return d*(n//d) + f(d-1, n)
and then I'd call the function with f(n, n).
I am trying this:
(defn f
([n] (f n n))
([d n]
(if (> d 0)
(recur (dec d) n)
0)))
But I don't know if this is right so far or where to slip in the sum or how to do it, etc.
If you look at your Clojure f function, the [d n] arity recurs with
d decremented and
n unchanged
... until d is zero, when it returns 0.
If we write this arity as a distinct local function, using letfn, we can drop the unchanging n argument, picking it up from the f argument:
(defn f [n]
(letfn [(g [d]
(if (> d 0)
(recur (dec d))
0))]
(g n)))
This produces the wrong answer of course, always returning 0:
(f 10)
=> 0
But we can see where to put the sum in:
(defn f [n]
(letfn [(g [d]
(if (> d 0)
(+ (* d (quot n d)) (g (dec d)))
0))]
(g n)))
We have to revert the recur to an explicit recursive call to g, as it is surrounded by the +.
But at least it works:
(f 10)
=> 87
In Clojure I want to make this a recursive function.
Don't. I've done it above just to show you where the calculation fits in.
Explicit recursion is rare in idiomatic Clojure. Better use the functions that encapsulate its common patterns. I won't repeat what Carciginate has given, but once you get used to threading macros, I think you'll find the following clear and concise:
(defn f [n]
(->> (range 1 (inc n))
(map (fn [d] (* d (quot n d))))
(reduce +)))
By the way, a reasonable analogue of your Python code is
(defn f [n]
(loop [s 0, d 1]
(if (> d n)
s
(recur (+ s (* d (quot n d))) (inc d)))))
I managed to get 3 ways working. Unfortunately, this algorithm doesn't seem to lend itself to nice recursion.
To get safe recursion, I had to introduce a third parameter. I just couldn't get it arranged so the recur was in the tail position. I also decided to count up instead of down. I don't think there's anything left field here, although it did get quite long unfortunately.
(defn f3
([n] (f3 n 1 0))
([n d s]
(if (> d (inc n))
s
(recur n (inc d)
(+ s (* d (quot n d)))))))
(f3 10)
If unsafe recursion is ok, this can be simplified quite a bit. Instead of adding multiple argument lists, I decided to allow d to be defaultable using & [d?]] and a check later down. I tend to avoid adding multiple argument lists since par-infer has a difficult time handling the indentation required to make it work. This trick isn't possible with the first way due to how recur handles var args. It only works if you're not using recur, or you do use recur, but only destructure 1 var-arg.
(defn f2 [n & [d?]]
(let [d (or d? 1)]
(if (> d (inc n))
0
(+ (f2 n (inc d)) (* d (quot n d))))))
(f2 10)
Unless you really need recursion though, I'd just write it as a map and reduction:
(defn f1 [n]
(reduce + 0
(map #(* % (quot n %)))
(range 1 (inc n)))))
(f1 10)
Which to me is about as neat as it gets (without using a threading macro. See Thumbnail's answer).
Try this:
(defn f
([n] (f n n))
([d n]
(if (> d 0)
(+ (* d (quot n d)) (recur (dec d) n))
0)))
Here is what I have done so far:
(define sumOdd
(lambda(n)
(cond((> n 0)1)
((odd? n) (* (sumOdd n (-(* 2 n) 1)
output would look something like this:
(sumOdd 1) ==> 1
(sumOdd 4) ==> 1 + 3 + 5 + 7 ==> 16
(sumOdd 5) ==> 1 + 3 + 5 + 7 + 9 ==> 25
This is what I am trying to get it to do: find the sum of the first N odd positive integers
I can not think of a way to only add the odd numbers.
To elaborate further on the sum-odds problem, you might solve it in terms of more abstract procedures that in combination accumulates the desired answer. This isn't necessarily the easiest solution, but it is interesting and captures some more general patterns that are common when processing list structures:
; the list of integers from n to m
(define (make-numbers n m)
(if (= n m) (list n) ; the sequence m..m is (m)
(cons n ; accumulate n to
(make-numbers (+ n 1) m)))) ; the sequence n+1..m
; the list of items satisfying predicate
(define (filter pred lst)
(if (null? lst) '() ; nothing filtered is nothing
(if (pred (car lst)) ; (car lst) is satisfactory
(cons (car lst) ; accumulate item (car lst)
(filter pred (cdr lst))) ; to the filtering of rest
(filter pred (cdr lst))))) ; skip item (car lst)
; the result of combining list items with procedure
(define (build-value proc base lst)
(if (null? lst) base ; building nothing is the base
(proc (car lst) ; apply procedure to (car lst)
(build-value proc base (cdr lst))))) ; and to the building of rest
; the sum of n first odds
(define (sum-odds n)
(if (negative? n) #f ; negatives aren't defined
(build-value + ; build values with +
0 ; build with 0 in base case
(filter odd? ; filter out even numbers
(make-numbers 1 n))))) ; make numbers 1..n
Hope this answer was interesting and not too confusing.
Let's think about a couple of cases:
1) What should (sumOdd 5) return? Well, it should return 5 + 3 + 1 = 9.
2) What should (sumOdd 6) return? Well, that also returns 5 + 3 + 1 = 9.
Now, we can write this algorithm a lot of ways, but here's one way I've decided to think about it:
We're going to write a recursive function, starting at n, and counting down. If n is odd, we want to add n to our running total, and then count down by 2. Why am I counting down by 2? Because if n is odd, n - 2 is also odd. Otherwise, if n is even, I do not want to add anything. I want to make sure that I keep recursing, however, so that I get to an odd number. How do I get to the next odd number, counting down from an even number? I subtract 1. And I do this, counting down until n is <= 0. I do not want to add anything to my running total then, so I return 0. Here is what that algorithm looks like:
(define sumOdd
(lambda (n)
(cond ((<= n 0) 0)
((odd? n) (+ n (sumOdd (- n 2))))
(else (sumOdd (- n 1))))))
If it helps you, here is a more explicit example of a slightly different algorithm:
(define sumOdd
(lambda (n)
(cond ((<= n 0) 0)
((odd? n) (+ n (sumOdd (- n 1))))
((even? n) (+ 0 (sumOdd (- n 1))))))) ; note that (even? n) can be replaced by `else' (if its not odd, it is even), and that (+ 0 ..) can also be left out
EDIT:
I see that the problem has changed just a bit. To sum the first N positive odd integers, there are a couple of options.
First option: Math!
(define sumOdd (lambda (n) (* n n)))
Second option: Recursion. There are lots of ways to accomplish this. You could generate a list of 2*n and use the procedures above, for example.
You need to have 2 variables, one which keep counter of how many odd numbers are still to be added and another to hold the current odd number which gets increment by 2 after being used in addition:
(define (sum-odd n)
(define (proc current start)
(if (= current 0)
0
(+ start (proc (- current 1) (+ start 2)) )))
(proc n 1))
Here is a nice tail recursive implementation:
(define (sumOdd n)
(let summing ((total 0) (count 0) (next 1))
(cond ((= count n) total)
((odd? next) (summing (+ total next)
(+ count 1)
(+ next 1)))
(else (summing total count (+ next 1))))))
Even shorter tail-recursive version:
(define (sumOdd n)
(let loop ((sum 0) (n n) (val 1))
(if (= n 0)
sum
(loop (+ sum val) (- n 1) (+ val 2)))))
I have the following 2 functions that I wish to combine into one:
(defun fib (n)
(if (= n 0) 0 (fib-r n 0 1)))
(defun fib-r (n a b)
(if (= n 1) b (fib-r (- n 1) b (+ a b))))
I would like to have just one function, so I tried something like this:
(defun fib (n)
(let ((f0 (lambda (n) (if (= n 0) 0 (funcall f1 n 0 1))))
(f1 (lambda (a b n) (if (= n 1) b (funcall f1 (- n 1) b (+ a b))))))
(funcall f0 n)))
however this is not working. The exact error is *** - IF: variable F1 has no value
I'm a beginner as far as LISP goes, so I'd appreciate a clear answer to the following question: how do you write a recursive lambda function in lisp?
Thanks.
LET conceptually binds the variables at the same time, using the same enclosing environment to evaluate the expressions. Use LABELS instead, that also binds the symbols f0 and f1 in the function namespace:
(defun fib (n)
(labels ((f0 (n) (if (= n 0) 0 (f1 n 0 1)))
(f1 (a b n) (if (= n 1) b (f1 (- n 1) b (+ a b)))))
(f0 n)))
You can use Graham's alambda as an alternative to labels:
(defun fib (n)
(funcall (alambda (n a b)
(cond ((= n 0) 0)
((= n 1) b)
(t (self (- n 1) b (+ a b)))))
n 0 1))
Or... you could look at the problem a bit differently: Use Norvig's defun-memo macro (automatic memoization), and a non-tail-recursive version of fib, to define a fib function that doesn't even need a helper function, more directly expresses the mathematical description of the fib sequence, and (I think) is at least as efficient as the tail recursive version, and after multiple calls, becomes even more efficient than the tail-recursive version.
(defun-memo fib (n)
(cond ((= n 0) 0)
((= n 1) 1)
(t (+ (fib (- n 1))
(fib (- n 2))))))
You can try something like this as well
(defun fib-r (n &optional (a 0) (b 1) )
(cond
((= n 0) 0)
((= n 1) b)
(T (fib-r (- n 1) b (+ a b)))))
Pros: You don't have to build a wrapper function. Cond constructt takes care of if-then-elseif scenarios. You call this on REPL as (fib-r 10) => 55
Cons: If user supplies values to a and b, and if these values are not 0 and 1, you wont get correct answer
My solution to exercise 1.11 of SICP is:
(define (f n)
(if (< n 3)
n
(+ (f (- n 1)) (* 2 (f (- n 2))) (* 3 (f (- n 3))))
))
As expected, a evaluation such as (f 100) takes a long time. I was wondering if there was a way to improve this code (without foregoing the recursion), and/or take advantage of multi-core box. I am using 'mit-scheme'.
The exercise tells you to write two functions, one that computes f "by means of a recursive process", and another that computes f "by means of an iterative process". You did the recursive one. Since this function is very similar to the fib function given in the examples of the section you linked to, you should be able to figure this out by looking at the recursive and iterative examples of the fib function:
; Recursive
(define (fib n)
(cond ((= n 0) 0)
((= n 1) 1)
(else (+ (fib (- n 1))
(fib (- n 2))))))
; Iterative
(define (fib n)
(fib-iter 1 0 n))
(define (fib-iter a b count)
(if (= count 0)
b
(fib-iter (+ a b) a (- count 1))))
In this case you would define an f-iter function which would take a, b, and c arguments as well as a count argument.
Here is the f-iter function. Notice the similarity to fib-iter:
(define (f-iter a b c count)
(if (= count 0)
c
(f-iter (+ a (* 2 b) (* 3 c)) a b (- count 1))))
And through a little trial and error, I found that a, b, and c should be initialized to 2, 1, and 0 respectively, which also follows the pattern of the fib function initializing a and b to 1 and 0. So f looks like this:
(define (f n)
(f-iter 2 1 0 n))
Note: f-iter is still a recursive function but because of the way Scheme works, it runs as an iterative process and runs in O(n) time and O(1) space, unlike your code which is not only a recursive function but a recursive process. I believe this is what the author of Exercise 1.1 was looking for.
I'm not sure how best to code it in Scheme, but a common technique to improve speed on something like this would be to use memoization. In a nutshell, the idea is to cache the result of f(p) (possibly for every p seen, or possibly the last n values) so that next time you call f(p), the saved result is returned, rather than being recalculated. In general, the cache would be a map from a tuple (representing the input arguments) to the return type.
Well, if you ask me, think like a mathematician. I can't read scheme, but if you're coding a Fibonacci function, instead of defining it recursively, solve the recurrence and define it with a closed form. For the Fibonacci sequence, the closed form can be found here for example. That'll be MUCH faster.
edit: oops, didn't see that you said forgoing getting rid of the recursion. In that case, your options are much more limited.
See this article for a good tutorial on developing a fast Fibonacci function with functional programming. It uses Common LISP, which is slightly different from Scheme in some aspects, but you should be able to get by with it. Your implementation is equivalent to the bogo-fig function near the top of the file.
To put it another way:
To get tail recursion, the recursive call has to be the very last thing the procedure does.
Your recursive calls are embedded within the * and + expressions, so they are not tail calls (since the * and + are evaluated after the recursive call.)
Jeremy Ruten's version of f-iter is tail-recursive rather than iterative (i.e. it looks like a recursive procedure but is as efficient as the iterative equivalent.)
However you can make the iteration explicit:
(define (f n)
(let iter
((a 2) (b 1) (c 0) (count n))
(if (<= count 0)
c
(iter (+ a (* 2 b) (* 3 c)) a b (- count 1)))))
or
(define (f n)
(do
((a 2 (+ a (* 2 b) (* 3 c)))
(b 1 a)
(c 0 b)
(count n (- count 1)))
((<= count 0) c)))
That particular exercise can be solved by using tail recursion - instead of waiting for each recursive call to return (as is the case in the straightforward solution you present), you can accumulate the answer in a parameter, in such a way that the recursion behaves exactly the same as an iteration in terms of the space it consumes. For instance:
(define (f n)
(define (iter a b c count)
(if (zero? count)
c
(iter (+ a (* 2 b) (* 3 c))
a
b
(- count 1))))
(if (< n 3)
n
(iter 2 1 0 n)))